Answer:
After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec
Explanation:
Area of a circle = πr²
where;
r is the circle radius
Differentiate the area with respect to time.

dr/dt = 4 ft/sec
after 12 seconds, the radius becomes = 
To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt


dA/dt = 1206.528 ft²/sec
Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec
Answer:
Explanation:
Here image distance is fixed .
In the first case if v be image distance
1 / v - 1 / -25 = 1 / .05
1 / v = 1 / .05 - 1 / 25
= 20 - .04 = 19.96
v = .0501 m = 5.01 cm
In the second case
u = 4 ,
1 / v - 1 / - 4 = 1 / .05
1 / v = 20 - 1 / 4 = 19.75
v = .0506 = 5.06 cm
So lens must be moved forward by 5.06 - 5.01 = .05 cm ( away from film )
Answer:
9ms^2
Explanation:
since ,Force=mass*acceleration
then, acceleration=force/mass
and, Force=90N
Mass=10pound
therefore, acceleration=90/10
=9ms^2
Answer : The power absorbed by the bulb is, 0.600 W
Explanation :
As we know that,
Power = Voltage × Current
Given:
Voltage = 3 V
Current = 200 mA = 0.200 A
Conversion used : (1 mA = 0.001 A)
Now put all the given values in the above formula, we get:
Power = Voltage × Current
Power = 3V × 0.200 A
Power = 0.600 W
Thus, the power absorbed by the bulb is, 0.600 W
Answer:
1.01 × 1013 picometres
Explanation:
multiply the length value by 1e+9